POJ 2318 TOYS(几何)

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题目链接：POJ 2318 TOYS

用点积判断一个点在直线的哪一侧，由于题目给出的分割线是排序后的，那么可以直接用二分得出答案。

#include <iostream>#include <cstdio>#include <cmath>#include <cstring>using namespace std;const double eps = 1e-10;const int MAX_N = 5000 + 10;struct Point{double x, y;Point(double x=0, double y=0):x(x),y(y) { }};typedef Point Vector;Vector operator + (const Vector& A, const Vector& B){    return Vector(A.x+B.x, A.y+B.y);}Vector operator - (const Point& A, const Point& B){    return Vector(A.x-B.x, A.y-B.y);}Vector operator * (const Vector& A, double p){    return Vector(A.x*p, A.y*p);}bool operator < (const Point& a, const Point& b){return a.x < b.x || (a.x == b.x && a.y < b.y);}int dcmp(double x){    if(fabs(x) < eps)        return 0;    else        return x < 0 ? -1 : 1;}bool operator == (const Point& a, const Point &b){return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;}double Cross(const Vector& A, const Vector& B){    return A.x*B.y - A.y*B.x;}Point p[MAX_N << 1];Point bound[3];Point toy;int _count[MAX_N];int n, m;void b_search(){    int l = 0, r = n - 1, mid;    while(l <= r)    {        mid = (l + r) >> 1;        if(Cross(toy - p[mid], p[mid + n] - p[mid]) > 0)            r = mid - 1;        else if(Cross(toy - p[mid], p[mid + n] - p[mid]) < 0)            l = mid + 1;    }    _count[r + 1]++;}int main(){    //freopen("in.txt", "r", stdin);    while(scanf("%d", &n), n)    {        memset(_count, 0, sizeof(_count));        scanf("%d", &m);        scanf("%lf%lf%lf%lf", &bound[0].x, &bound[0].y, &bound[2].x, &bound[2].y);        for(int i = 0; i < n; i++)        {            scanf("%lf%lf",&p[i].x, &p[i + n].x);            p[i].y = bound[0].y;            p[i + n].y = bound[2].y;        }        for(int i = 0; i < m; i++)        {            scanf("%lf%lf", &toy.x, &toy.y);            b_search();        }        for(int i = 0; i <= n; i++)            printf("%d: %d\n", i, _count[i]);        printf("\n");    }    return 0;}

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